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傳送門：

http://acm.hdu.edu.cn/showproblem.php?pid=1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 63806 Accepted Submission(s): 27457

Problem Description A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

Input n (0 < n < 20).

Output The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

Sample Input 6 8

Sample Output Case 1: 1 4 3 2 5 6 1 6 5 2 3 4 Case 2: 1 2 3 8 5 6 7 4 1 2 5 8 3 4 7 6 1 4 7 6 5 8 3 2 1 6 7 4 3 8 5 2 Source Asia 1996, Shanghai (Mainland China)

Recommend JGShining | We have carefully selected several similar problems for you: 1312 1072 1242 1175 1253 分析： 經典的素數環問題 dfs 就是全排列的基礎上加上素數環要求的檢測 code：

#include<bits/stdc++.h>
using
 
 namespace std;
typedef long long LL;
#define max_v 105
int a[max_v];
int vis[max_v];
int n;
int isp(int x)//素數檢測
{
    for(int i=2;i<=sqrt(x);i++)
    {
        if(x%i==0)
            return 0;
    }
    return 1;
}
void dfs(int cur)//素數環問題 全排列思想加素數的檢測
{
    if(cur==n&&isp(a[0]+a[n-1]))//判斷到最後一個數了
    {
        printf("%d",a[0]);//打印
        for(int i=1;i<n;i++)
        {
            printf(" %d",a[i]);
        }
        printf("\n");
        return ;
    }else
    {
        for(int i=2;i<=n;i++)//找適合放在cur位置的i
        {
            if(!vis[i]&&isp(i+a[cur-1]))//滿足要求
            {
                a[cur]=i;//放入
                vis[i]=1;//標記
                dfs(cur+1);//搜索
                vis[i]=0;//回退
            }
        }
    }
}
int main()
{
    int t=1,k=0;
    while(~scanf("%d",&n))
    {
        //if(k)
            //printf("\n");
        memset(vis,0,sizeof(vis));//標記清空
        a[0]=1;//確定1的位置
        printf("Case %d:\n",t);
        dfs(1);//從1開始放數
        t++;
        k++;
        printf("\n");
    }
    return 0;
}

HDU 1016 Prime Ring Problem（素數環問題）